Question Number 134327 by mr W last updated on 02/Mar/21 | ||
Answered by aleks041103 last updated on 25/Dec/21 | ||
$${i}\Rightarrow\mid{z}_{{k}} ^{\mathrm{2}} \mid=\mathrm{1} \\ $$$$\Rightarrow{z}_{{k}} ={e}^{{it}_{{k}} } \\ $$$${w}_{{k}} ={z}_{{k}} ^{\mathrm{2}} ={e}^{\mathrm{2}{it}_{{k}} } \\ $$$$\Rightarrow{w}_{\mathrm{1}} +{w}_{\mathrm{2}} +{w}_{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}+{e}^{\mathrm{2}{i}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)} +{e}^{\mathrm{2}{i}\left({t}_{\mathrm{3}} −{t}_{\mathrm{1}} \right)} =\mathrm{0} \\ $$$$\Rightarrow{Im}\left({e}^{\mathrm{2}{i}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)} \right)=−{Im}\left({e}^{\mathrm{2}{i}\left({t}_{\mathrm{3}} −{t}_{\mathrm{1}} \right)} \right) \\ $$$$\Rightarrow{t}_{\mathrm{2}} −{t}_{\mathrm{1}} ={t}_{\mathrm{1}} −{t}_{\mathrm{3}} \\ $$$${also} \\ $$$$\mathrm{2}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)=\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2}\pi{m} \\ $$$$\Rightarrow{t}_{\mathrm{2}} −{t}_{\mathrm{1}} =\frac{\pi}{\mathrm{3}}+{m}\pi \\ $$$$\Rightarrow{t}_{\mathrm{3}} −{t}_{\mathrm{1}} ={p}\pi−\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{z}_{\mathrm{2}} ={z}_{\mathrm{1}} {e}^{{i}\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)} ={z}_{\mathrm{1}} {e}^{{i}\frac{\pi}{\mathrm{3}}} {e}^{{im}\pi} =\pm{z}_{\mathrm{1}} {e}^{\frac{{i}\pi}{\mathrm{3}}} \\ $$$${z}_{\mathrm{3}} ={z}_{\mathrm{1}} {e}^{{i}\left({t}_{\mathrm{3}} −{t}_{\mathrm{1}} \right)} ={z}_{\mathrm{1}} {e}^{−\frac{{i}\pi}{\mathrm{3}}} {e}^{{ip}\pi} =\pm{z}_{\mathrm{1}} {e}^{−\frac{{i}\pi}{\mathrm{3}}} \\ $$$${z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} = \\ $$$$={z}_{\mathrm{1}} \left(\mathrm{1}\pm{e}^{\frac{{i}\pi}{\mathrm{3}}} \pm{e}^{−\frac{{i}\pi}{\mathrm{3}}} \right)\neq\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}+{s}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+{s}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\neq\mathrm{0} \\ $$$${if}\:−\mathrm{2}={s}_{\mathrm{1}} +{s}_{\mathrm{2}} \:{and}\:{s}_{\mathrm{1}} −{s}_{\mathrm{2}} =\mathrm{0}\:{then}\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} =\mathrm{0} \\ $$$${only}\:{when}\:{s}_{\mathrm{1}} ={s}_{\mathrm{2}} =−\mathrm{1} \\ $$$$ \\ $$$${Case}\:\mathrm{1}: \\ $$$${z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} =−{z}_{\mathrm{1}} {e}^{{i}\pi/\mathrm{3}} ={z}_{\mathrm{1}} {e}^{\mathrm{4}{i}\pi/\mathrm{3}} ,\:{z}_{\mathrm{3}} ={z}_{\mathrm{1}} {e}^{−{i}\pi/\mathrm{3}} \\ $$$$\mid{z}_{\mathrm{1}} ^{{n}} +{z}_{\mathrm{2}} ^{{n}} +{z}_{\mathrm{3}} ^{{n}} \mid=\mid{z}_{\mathrm{1}} \mid^{{n}} \:\mid\mathrm{1}+{e}^{\mathrm{4}{ni}\pi/\mathrm{3}} +{e}^{−{ni}\pi/\mathrm{3}} \mid= \\ $$$$=\mid\mathrm{1}+{e}^{\mathrm{4}{ni}\pi/\mathrm{3}} +{e}^{−{ni}\pi/\mathrm{3}} \mid \\ $$$$.... \\ $$$${the}\:{rest}\:{is}\:{trivial} \\ $$ | ||